Graph this system of equations and solve. $y = -\dfrac{6}{5} x - 1$ $20x+10y = -50$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: The y-intercept for the first equation is $-1$ , so the first line must pass through the point $(0, -1)$ The slope for the first equation is $-\dfrac{6}{5}$ . Remember that the slope tells you rise over run. So in this case for every $6$ positions you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. $6$ positions down from $(0, -1)$ is $(5, -7)$ Graph the blue line so it passes through $(0, -1)$ and $(5, -7)$ Convert the second equation, $20x+10y = -50$ , to slope-intercept form. $y = -2 x - 5$ The y-intercept for the second equation is $-5$ , so the second line must pass through the point $(0, -5)$ The slope for the second equation is $-2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move down (because it's negative) You must also move $1$ positions to the right. $1$ position to the right. $2$ positions down from $(0, -5)$ is $(1, -7)$ Graph the green line so it passes through $(0, -5)$ and $(1, -7)$ The solution is the point where the two lines intersect. The lines intersect at $(-5, 5)$.